Question

The average kinetic energy of one molecule of an ideal gas at ${{27}^{\circ }}C$ and 1atm pressure is:
[A] $900\text{ cal }{{\text{K}}^{-1}}mo{{l}^{-1}}$
[B] $6.21\times {{10}^{-21}}J\text{ }{{K}^{-1}}molecule{{e}^{-1}}$
[C] $336.7J\text{ }{{K}^{-1}}molecule{{e}^{-1}}$
[D] $3741.3\text{ J }{{\text{K}}^{-1}}mo{{l}^{-1}}$

Answer 1

HINT: The average kinetic energy if dependent only upon the temperature. You can solve this by using the formula- $K.E=\frac{3}{2}{{K}_{B}}T$, where the terms have their usual meanings. Do not forget that the Boltzmann constant can be written as- ${{K}_{B}}=\frac{R}{{{N}_{A}}}$.

COMPLETE STEP BY STEP SOLUTION: We know that the kinetic energy of a particle is the energy that it possesses due to motion. We define it as the work needed to accelerate a body of a certain given mass from rest.
According to the kinetic theory of gases, we can find the average kinetic energy of per molecule any particle by the formula-
     $K.E=\frac{3}{2}{{K}_{B}}T$
Where, ${{K}_{B}}$ is Boltzmann's constant, T is the temperature and K.E is the average kinetic energy.
We can also rewrite the above formula for per mole of a gas as-
     $K.E=\frac{3}{2}nRT$
Where, n is the number of moles of the particular gas, T is the temperature, R is the universal gas constant whose value is fixed and K.E is the average kinetic energy.
Now, for one molecule of ideal gas, we can write the first formula as-
     $K.E=\frac{3}{2}\frac{RT}{{{N}_{A}}}$
We know that Boltzmann’s constant ${{K}_{B}}=\frac{R}{{{N}_{A}}}$
Where, ${{N}_{A}}$ is Avogadro's number.
Now in the question the temperature is given as ${{27}^{\circ }}C$ but we have to convert it in kelvin. We know that ${{0}^{\circ }}C$ is equal to 273 K.
Therefore, ${{27}^{\circ }}C$ is 27 + 273 K = 300K.
The value of the universal gas constant, R = 8.314 J/Kmol
And we know that the value of Avogadro’s number, ${{N}_{A}}$= $6.022\times {{10}^{23}}$
Now we will put these values in the average kinetic energy equation to find out its value for one molecule of an ideal gas.
     $K.E=\frac{3}{2}\times \frac{8.314\text{ }J/Kmol\times 300K}{6.022\times {{10}^{23}}}=6.21\times {{10}^{-21}}J/molecule$
We can see from the above calculation that the value of average kinetic energy for one molecule of an ideal gas is $6.21\times {{10}^{-21}}J/molecule$.


Therefore the correct answer is option [B] $6.21\times {{10}^{-21}}J\text{ }{{K}^{-1}}molecule{{e}^{-1}}$

NOTE: The average kinetic energy of a particle is dependent only upon the absolute temperature of the system. The absolute temperature is a scale for measurement of temperature of an object where 0 is taken as absolute zero. The absolute temperature scales are kelvin and Rankine.